so,we have nothing to start with ? lets rearrange L.H.S and see if we find something useful.

given L.H.S is (3a+2007)^{2}

taking 3 common.

[3(a+669)]^{2}

9(a+669)^{2}. now write lhs=rhs.

9(a+669)^{2}=4b85924--eq(1) after analyzing this equation we reach to the conclusion that 9 is a factor of number on RHS.

Now we can use divisiblity criteria of 9 to move ahead.

what is divisiblity criteria of 9?... check wikis for that

《USING DIVISIBLITY CRITERIA OF 9》 now since 4b85924 is multiple of 9.therefore sum of its digits must be a multiple of 9,also keep in mind that b is a digit. NoW, if K is sum of digits of 4b85924 then K=4+b+8+5+9+2+4. or K=b+32--eq(2).

now this K in (eq.2) must be a multiple of 9,as sum of digits(here K) of a multiple of 9(here 4b85924) must be a multiple of 9.. after putting values [b=1,2,3...9] one by one in eq2 and analyzing eq2 after putting values.

we noticed that b=4 is appropiate value of b in eq(2). as K=4+32=36 is a multiple of 9[also note that b is a digit so u can put only 0,1,2..9 as value of b].

now putting value of b in eq--(1). (a+669)^{2}=4485924/9. => (a+669)=706. OR (a+669)=-706. which gives a=-1375 or 37.

since,a is a natural number(given in question) therfore we have to take a=37 now a-b=¿.

a-b=37-4 =33 so answer is 33. :)